Suppose $4$ balls labeled $1,2,3,4$ are randomly placed in boxes $B_1,B_2,B_3,B_4$. The probability that exactly one box is empty is
(a)$8/256$
(b)$9/16$
(c)$27/256$
(d)$9/64$
My approach:
Selecting one empty box out of the four boxes, then placing the $4$ balls in the remaining $3$ boxes.
Thank you .
Best Answer
Let's implement your strategy.
There are $\binom{4}{1}$ ways to choose the empty box. We must distribute the four balls to the remaining three boxes so that no box is left empty. That means one of those three boxes will receive two balls and the others will each receive one. There are $\binom{3}{1}$ ways to choose the box that will receive two balls and $\binom{4}{2}$ ways to choose which two balls will be placed in that box. The remaining balls can be placed in the remaining two boxes in $2!$ ways. Hence, the number of favorable cases is $$\binom{4}{1}\binom{3}{1}\binom{4}{2}2!$$ Since there are $4$ choices for the placement of each of the four balls, there are $4^4$ ways to distribute four distinct balls to four distinct boxes. Hence, the desired probability is $$\frac{\dbinom{4}{1}\dbinom{3}{1}\dbinom{4}{2}2!}{4^4}$$